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1.

Let E and F be two independent events. The probability that exactly one of them occurs is `11//25` and the probability of none of them occurring is `2//25.` If P(T) denotes the probability of occurrence of the event T, thenA. `P(E)=4/5,P(F)=3/5`B. `P(E)=1/5,P(F)=2/5`C. `P(E)=2/5,P(F)=1/5`D. `P(E)=3/5,P(F)=4/5`

Answer» Correct Answer - A::D
Let `P(E)=eand P(F)=f`
`P(EuuF)-P(EnnF)=11/25`
`impliese+f-2ef=11/25" "(1)`
`P(barEnnbarF)=2/25`
`implies(1-e)(1-f)=2/25" "(2)`
From (1) and (2),
`ef=12/25and e+f=7/5`
Solving, we get
`e=4/5,f=3/5or e=3/5,f=4/5`
Discussion
2.

Of the three independent event `E_(1),E_(2)` and `E_(3)`, the probability that only `E_(1)` occurs is `alpha`, only `E_(2)` occurs is `beta` and only `E_(3)` occurs is `gamma`. If the probavvility p that none of events `E_(1), E_(2)` or `E_(3)` occurs satisfy the equations `(alpha - 2beta)p = alpha beta` and `(beta - 3 gamma) p = 2 beta gamma`. All the given probabilities are assumed to lie in the interval (0, 1). Then, `("probability of occurrence of " E_(1))/("probability of occurrence of " E_(3))` is equal to

Answer» Let `P(E_(1))=x,P(E_(2))=y and P(E_(3))=z,then`
`(1-x)(1-y)(1-z)=p`
`x(1-y)(1-z)=alpha,`
`(1-x)y(1-z)=beta,`
`(1-x)(1-y)(z)=gamma`
`So, (1-x)/(x)=(p)/(alpha)=x=(alpha)/(alpha+p),`
Similarly, `z=(gamma)/(gamma+p).`
So, `(P(E_(1)))/(P(E_(3)))=((alpha)/(alpha+p))/(gamma/gamma+p)=((gamma+p)/(gamma))/((alpha+p)/(alpha))=(a+p/gamma)/(1+(p)/(alpha))`
Also given `(alphabeta)/(alpha-2beta)=p=(2betagamma)/(beta=3gamma)impliesbeta=(5alphagamma)/(alpha+4gamma)`
Substituting in given relation
`(alpha-2((5alpha gamma)/(alpha+4gamma)))p=(alphaxx5alphagamma)/(alpha+4gamma)`
`or ((p)/(gamma)+1)=6((p)/(alpha)+1)`
`or ((p)/(gamma+1))/((p)/(alpha)+1)=6.`
Discussion
3.

Football teams `T_(1)and T_(2)` have to play two games are independent. The probabilities of `T_(1)` winning, drawing and lossing a game against `T_(2)` are `1/6,1/6and 1/3,` respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams `T_(1) and T_(2)` respectively, after two games. `P(X=Y)` isA. `11/36`B. `1/3`C. `13/36`D. `1/2`

Answer» Correct Answer - C
`P(X=Y)=DD+T_(1)T_(2)+T_(2)T_(1)`
`=((1)/(6)xx(1)/(6))+((1)/(2)xx(1)/(3))+((1)/(3)xx(1)/(2))=13/36`
Discussion
4.

A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses required. The probability that `X=3` equalsA. `25//216`B. `25//36`C. `5//36`D. `125//216`

Answer» Correct Answer - A
`P(X=3)=((5)/(6))((5)/(6))1/6=25/216`
Discussion
5.

Three numbers are chosen at random without replacement from `{1,2,3,...,8}.` The probability that their minimum is 3, given that maximum is 6, is:A. `3/8`B. `1/5`C. `1/4`D. `2/5`

Answer» Correct Answer - B
Given set of numbers is `{1,2,3,4,5,6,7,8}`
We have to select three numbers
Let us assume that
event A `to` inselected three numbers minimum is 3
event B `to` in selected three numbers maximum is 6
We have to find P (A/B).
`P(A//B)=(P(AnnB))/(P(B))=(n(AnnB))/(N(B))`
`AnnB` = in selected three numbers, minimum is 3 and maximum is 6
`therefore` The remaining number is either 4 or 5
`therefore n(AnnB)=2`
n(B) = selecting two numbers from `1,2,3,4,5` (as maximum of numbers is 6)
`=""^(5)C_(2)`
`thereforeP(A//B)=(n(AnnB))/(n(B))=(2)/(""^(5)C_(2))=1/5`
Discussion
6.

If C and D are two events such that `C subD`and P(D) is not equal to 0,` then the correct statement among the following isA. `P(C|D)=(P(D))/(P(D))`B. `P(C|D)=P(C)`C. `P(C|D)geP(C)`D. `P(C|D)ltP(C)`

Answer» Correct Answer - C
We have `CsubD`
`impliesCnnD=C`
`impliesP((C)/(D))=(P(CnnD))/(P(D))=(P(C))/(P(D))geP(C)`
As `0ltP(D)le1`
Discussion
7.

If A and B are two events such that `P(A)=0.6 and P(B)=0.8,` if the greatest value that `P(A//B)` can have is p, then the value of 8p is______.

Answer» Correct Answer - 6
`P(A//B)=(P(AnnB))/(P(B))=(0.6)/(0.8)=3/4`
`" "("Maximum value of"P(AnnB)=P(A)=0.6)`
Discussion
8.

One ticket is selected at random from 50 ticketsnumbered 00, 01, 02, ... , 49. Then the probability that the sum of thedigits on the selected ticket is 8, given that the product of these digits iszero, equals(1) 1/14 (2) 1/7(3) 5/14(4) 1/50A. `1/14`B. `1/7`C. `5/14`D. `1/50`

Answer» Correct Answer - A
`S={00,01,02,....,49}`
Let A be the event that the sum of the digits on the selected ticket is 8. Then
`A={08,17,26,35,44}`
Let B be the event that the product of the digits is zero. Than
`B={00,01,02,03,...,09,10,20,30,40}`
`thereforeAnnB={8}`
The required probability is
`P(A//B)=(P(AnnB))/(P(B))`
`=(1/50)/(14/50)=1/14`
Discussion
9.

An urn contains 3 red balls and `n`white balls. Mr. A draws two balls together from the urn. Theprobability that they have the same color is 1/2 Mr. B. Draws one balls formthe urn, notes its color and replaces it. He then draws a second ball fromthe urn and finds that both balls have the same color is 5/8. The possiblevalue of `n`is`9`b. `6`c. `5`d. 1

Answer» Correct Answer - 1
There are n white balls in the turn.
`implies` Probability of Mr. A to draw two balls of same color is
`(""^(3)C_(2)+""^(n)C_(2))/(""^(n+3)C_(2))=1/2("given")`
`or (6+n(n-1))/((n+3)(n+2))=1/2`
`or n^(2)-7n+6=0`
`impliesn=1or6" "(1)`
Also required probability for Mr. B according to the question is
`(3)/(n+3)(3)/(n+3)+(n)/(n+3)(n)/(n+3)=5/8"(given)"`
Solving, we get `n^(2)-10n +9=0,n=1or9" "(2)`
From (1) and (2), `n=1`
Discussion
10.

A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls and 4 black balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls. If 2 balls are drawn (without replecement) from a randomly selected box and one of the balls is white and the other ball is red the probability that these 2 balls are drawn from box `B_(2)` isA. `116//182`B. `126//181`C. `65//181`D. `55//181`

Answer» Correct Answer - D
Let A: one ball is white and other is red
`E_(1):` both balls are from box `B_(1),`
`E_(2): ` both balls are from box `B_(2),`
`E_(3):` both balls are form box `B_(3)`
Hence, P (required) `=P((E_(2))/(A))`
`=(P((A)/(E_(2))).P(E_(2)))/(P((A)/(E_(1))).P(E_(1))+P((A)/(E_(2))).P(E_(2))+P((A)/(E_(3))).P(E_(3)))`
`=((""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3)/((""^(1)C_(1)xx""^(3)C_(1))/(""^(6)C_(2))xx1/3+(""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3+(""^(3)C_(1)xx""^(4)C_(1))/(""^(12)C_(2))xx1/3)`
`(1/6)/(1/5+1/6+2/11)=55/181`
Discussion
11.

A single which can can be green or red with probability `4/5 and 1/5` respectively, is received by station A and then transmitted to station B. The probability of each station reciving the signal correctly is `3/4.` If the singal received at station B is green, then the probability that original singal was green isA. `3/5`B. `6/7`C. `20/23`D. `9/20`

Answer» Correct Answer - C
Even G = original signal is green
`E_(1)=A` receives the signal correct
`E_(2)=B` receives the signal correct
E = Signal received by B is green
P (Signal received by B is green)
`=P(GE_(1)E_(2))+P(GbarE_(1)barE_(2))+P(barGE_(1)barE_(2))+P(barGbarE_(1)E_(2))`
`P(E)=(46)/(5xx16)`
`P(G//E)(40//5xx16)/(46//5xx16)=20/23`
Discussion